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Let $\Omega = \{\; 0, 1 \;\}^{\mathbb{N}}$ be the set of all functions $\mathbb{N} \to \{\; 0, 1 \;\}$. Then $\displaystyle\mathcal{F} = \bigotimes_{n \in \mathbb{N}} \mathcal{P}(\{\; 0, 1 \;\})$ is a $\sigma$-algebra on $\Omega$, where $\mathcal{P}(\dots)$ denotes the power set and $\bigotimes$ the product-$\sigma$-algebra.

Is $\mathcal{F} = \mathcal{P}(\Omega)$? This seems unlikely, but how would one construct a non $\mathcal{F}$-measurable set? As $(\{\; 0, 1 \;\}, \mathcal{P}(\{\; 0, 1 \;\}))$ is a polish space, one can prove that $\mathcal{F}$ is the Borel-$\sigma$-algebra generated by the product topology on $\Omega$.

MeinAccount
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1 Answers1

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Let $f: \Omega \to C$ be a homeomorphism between $\Omega$ and a fat Cantor subset $C$ of $[0,1]$ (see here).

Since $C$ has positive Lebesgue measure, it has a non-measurable subset $A$. Then, $f^{-1}(A)$ is non-measurable.

For more details, see this question and answer.

aduh
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